Why the alternate form of $frac{sqrt3}2+frac i 2$ is $sqrt[6]{-1}$?












0












$begingroup$


The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46
















0












$begingroup$


The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46














0












0








0





$begingroup$


The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic







share|cite|improve this question











$endgroup$




The alternate form of:
$$frac{sqrt3}2+frac i 2$$
is
$$sqrt[6]{-1}$$
(I know that thanks to WolframAlpha.)



What are the arithmetic actions that gets us from the former to the latter?






algebra-precalculus complex-numbers arithmetic




algebra-precalculus complex-numbers arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:51









gimusi

1




1










asked Dec 13 '18 at 22:33









HeyJudeHeyJude

1636




1636








  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46














  • 2




    $begingroup$
    Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 22:35










  • $begingroup$
    The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
    $endgroup$
    – user
    Dec 13 '18 at 22:44












  • $begingroup$
    Related: "Principal $n$th root of a complex number".
    $endgroup$
    – Blue
    Dec 13 '18 at 22:46








2




2




$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35




$begingroup$
Well you can simply raise $sqrt{3}/2+i/2$ to the sixth power and find out that you get $-1$.
$endgroup$
– SmileyCraft
Dec 13 '18 at 22:35












$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44






$begingroup$
The two forms are not equivalent. $sqrt[6]{-1}$ is an ambiguous presentation of 6 different complex numbers.
$endgroup$
– user
Dec 13 '18 at 22:44














$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46




$begingroup$
Related: "Principal $n$th root of a complex number".
$endgroup$
– Blue
Dec 13 '18 at 22:46










3 Answers
3






active

oldest

votes


















3












$begingroup$

That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
$$
omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
$$
Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



    The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      We simply have that



      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



      Refer also to:




      • Principal $n$th root of a complex number






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
        $endgroup$
        – Blue
        Dec 13 '18 at 22:44












      • $begingroup$
        That's the principal root! Ok I add something more on that.
        $endgroup$
        – gimusi
        Dec 13 '18 at 22:46











      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038663%2fwhy-the-alternate-form-of-frac-sqrt32-frac-i-2-is-sqrt6-1%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



      The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



      The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
      $$
      omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
      $$
      Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



        The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



        The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
        $$
        omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
        $$
        Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



          The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



          The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
          $$
          omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
          $$
          Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.






          share|cite|improve this answer











          $endgroup$



          That's a terrible notation by Wolfram. There are six sixth roots of $-1$, so you cannot tell which one it is when you write $sqrt[6]{-1}$.



          The notation makes sense when we write $sqrt{2}$, say, because we take the usual convention that it is the positive root. But such choice is meaningless for arbitrary roots.



          The sixth roots of $-1$ are, from De Moivre's formula (and writing $-1=cospi+isin pi$),
          $$
          omega_k=cosleft(tfrac{pi+2kpi}{6}right)+isinleft(tfrac{pi+2kpi}{6}right), k=0,ldots,5.
          $$
          Your root is $omega_0$ above. It is what we call a primitive root, in the sense that 6 is the smallest positive integer $r$ such that $omega_0^r=-1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 14 '18 at 12:06









          Scientifica

          6,37641335




          6,37641335










          answered Dec 13 '18 at 22:41









          Martin ArgeramiMartin Argerami

          124k1177176




          124k1177176























              2












              $begingroup$

              $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



              The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



                The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



                  The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.






                  share|cite|improve this answer









                  $endgroup$



                  $$sqrt{3}/2+i/2=cos(pi/6)+isin(pi/6)=e^{ipi/6}"="(e^{ipi})^{1/6}=(-1)^{1/6}.$$



                  The second equality is Euler's theorem. The third equals sign is in quotes is because technically there are $6$ values of $(e^{ipi})^{1/6}$, and $e^{ipi/6}$ happens to be one of them.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 13 '18 at 22:39









                  Mike EarnestMike Earnest

                  20.9k11951




                  20.9k11951























                      0












                      $begingroup$

                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46
















                      0












                      $begingroup$

                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number






                      share|cite|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46














                      0












                      0








                      0





                      $begingroup$

                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number






                      share|cite|improve this answer











                      $endgroup$



                      We simply have that



                      $$left(frac{sqrt3}2+frac i 2right)^6=(e^{ipi/6})^6=e^{ipi}=-1$$



                      and since $frac{sqrt3}2+frac i 2$ is the principal root of $z^6+1=0$ for that reason it is designed as $sqrt[6]{-1}$.



                      Refer also to:




                      • Principal $n$th root of a complex number







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Dec 13 '18 at 22:49

























                      answered Dec 13 '18 at 22:37









                      gimusigimusi

                      1




                      1








                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46














                      • 1




                        $begingroup$
                        This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                        $endgroup$
                        – Blue
                        Dec 13 '18 at 22:44












                      • $begingroup$
                        That's the principal root! Ok I add something more on that.
                        $endgroup$
                        – gimusi
                        Dec 13 '18 at 22:46








                      1




                      1




                      $begingroup$
                      This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                      $endgroup$
                      – Blue
                      Dec 13 '18 at 22:44






                      $begingroup$
                      This demonstrates that $sqrt{3}/2+i/2$ is a sixth-root of $-1$, but it doesn't explain why the value deserves the designation "$sqrt[6]{-1}$".
                      $endgroup$
                      – Blue
                      Dec 13 '18 at 22:44














                      $begingroup$
                      That's the principal root! Ok I add something more on that.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 22:46




                      $begingroup$
                      That's the principal root! Ok I add something more on that.
                      $endgroup$
                      – gimusi
                      Dec 13 '18 at 22:46


















                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038663%2fwhy-the-alternate-form-of-frac-sqrt32-frac-i-2-is-sqrt6-1%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Bressuire

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more

                      Cabo Verde

                      Image
                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
                      Read more

                      Gyllenstierna

                      Image
                      .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
                      Read more